How can I find an equation of motion of this? There is a point of mass moving in the xy-plane with harmonic forces acting in x- and y-direction 
$F_x=-m\omega^2x$ and $F_y=-m\omega^2y$. At the same time there is an additional force acting in the x-direction with $F_x'=\alpha m\omega^2y$ ($\alpha$ > 0). 
I am supposed to find the equation of motion of this with the IVP: 
$x(0)=y(0)=0, ~\dot x(0)=0,~\dot y(0)=A\omega$
I was thinking of adding the forces to the total force and then separate for x and y each time and having the equation of motion of x in dependance of y and the total force and vice versa. But the teacher said that this was the wrong approach. Can someone help me out? 
He said that the general solution of the differential equation is of the form of $\ddot z=-\omega^2z$. 
 A: Let $\mathbf{s}=\begin{bmatrix}x\\y \end {bmatrix}$ the position of the point.
The force acting on it is $ \mathbf{F}=\begin{bmatrix}-m\omega^2x+\alpha m\omega^2y\\-m\omega^2y \end {bmatrix}$. So the equation of motion is:
$$
\ddot{\mathbf{s}}=\begin{bmatrix}\ddot x\\\ddot y \end {bmatrix}=
\begin{bmatrix}-\omega^2x+\alpha \omega^2y\\-\omega^2y \end {bmatrix}
$$
The $y$ component, with initial conditions $y(0)=0$ and $y'(0)=A\omega$, gives the classical solution for an harmonic oscillator: $y=A\sin(\omega t)$.
Substituting this solution in the $x$ component we have the equation:
$$
\ddot x+\omega^2 x=\alpha \omega^2\sin(\omega t)
$$
that is the equation of an harmonic oscillator with a forcing term, so its solution is of the form:
$$
x=c_1\cos(\omega t)+c_2\sin(\omega t) +F(t)
$$
where $F(t)$ is a particular solution.  We can find this particular solution with method of undetermined coefficients searching for a function of the form:
$ F(t)=t(k\sin(\omega t)+h\cos(\omega t))$. With a bit of calculus we find:
$$
F(t)=-\dfrac{1}{2}\left(\alpha \omega A t \cos(\omega t)  \right)
$$
so we have the solution:
$$
x(t)=c_1 \cos(\omega t)+ c_2 \sin (\omega t)-\dfrac{1}{2}\left(\alpha \omega A t \cos(\omega t)  \right)
$$
The initial condition $x(0)=0$ gives $c_1=0$ and from $x'(0)=0$ we  find $ c_2=\dfrac{1}{2}\alpha \omega A$, so the solution is:
$$
\mathbf{s}(t)=\begin{bmatrix}x(t)\\y(t) \end {bmatrix}=\begin{bmatrix}
 \dfrac{1}{2}\alpha \omega A\left( \sin (\omega t)-t\cos(\omega t)\right)\\A\sin(\omega t)\end {bmatrix}
$$
